Hướng làm: Áp dụng công thức \(A.B>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}A>0\\B>0\end{matrix}\right.\\\left\{{}\begin{matrix}A< 0\\B< 0\end{matrix}\right.\end{matrix}\right.\)
\(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x>2\) hoặc \(x< -\dfrac{2}{3}\)
Ta có: \(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{2}{3}\end{matrix}\right.\)